A parallel plate capacitor has the space between its plates filled by two slabs of thickness d2 each and dielectric constant k2 and d is the plate separation of the capacitor- The capacitance of the capacitor is-

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Standard XII

Physics

Polarization Vector

A parallel pl...

Question

A parallel plate capacitor has the space between its plates filled by two slabs of thickness $\frac{d}{2}$ each and dielectric constant $k_{2}$ and d is the plate separation of the capacitor. The capacitance of the capacitor is:

\frac{2 ε_{0} A}{d} (\frac{K_{1} + K_{2}}{K_{1} K_{2}})

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\frac{2 ε_{0} A}{d} (K_{1} + K_{2})

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\frac{2 ε_{0} A}{d} (\frac{K_{1} K_{2}}{K_{1} + K_{2}})

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\frac{2 ε_{0} d}{A} (\frac{K_{1} + K_{2}}{K_{1} K_{2}})

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Solution

The correct option is C $\frac{2 ε_{0} A}{d} (\frac{K_{1} K_{2}}{K_{1} + K_{2}})$
the above effective capacitance is series combination of $\frac{K_{1} ϵ_{0} A_{2}}{d}$ and $\frac{K_{2} ϵ_{0} A_{2}}{d}$
$∴ \frac{1}{C_{b f}} = \frac{d}{K_{1} ϵ_{0} A_{2}} + \frac{d}{K_{2} ϵ_{0} A_{2}}$
$\Rightarrow C_{b f} = \frac{2 ϵ_{0} A}{d} (\frac{K_{1} K_{2}}{K_{1} + K_{2}})$

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